Problem
Each move removes two coins from one pile and one coin from the other. Either orientation is allowed. You may never remove more coins than remain. Determine whether both piles can end at zero.
Implement coin_piles(left, right).
Input
left: The number of coins in the left pile.
right: The number of coins in the right pile.
Output
Return true if the piles can both be emptied; otherwise return false.
Constraints
0 ≤ left, right ≤ 1,000,000,000. Two empty piles are already a successful result.
Examples
Input:
left = 4
right = 2
Expected output:
expected = TrueInput:
long long left = 4;
long long right = 2;
Expected output:
bool expected = true;Input:
long left = 4;
long right = 2;
Expected output:
bool expected = true;Explanation
Remove (2, 1) twice: (4, 2) → (2, 1) → (0, 0).
Input:
left = 4
right = 1
Expected output:
expected = FalseInput:
long long left = 4;
long long right = 1;
Expected output:
bool expected = false;Input:
long left = 4;
long right = 1;
Expected output:
bool expected = false;Explanation
The only possible first move leaves (2, 0), after which no move can empty the left pile.
How to approach it
Try on your own first, then open one hint at a time.
Hint 1
Each move removes exactly three coins total, so the overall coin count must fit that pattern.
Hint 2
Neither pile can be reduced below zero, so one pile cannot exceed what the other can support.
Hint 3
Check both the total amount and the difference between piles.
Expected complexity
O(1) time, O(1) extra space
References
This challenge is adapted for CodeUtility. Follow the statement and constraints above.